Answer:
[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]
Explanation:
As we know by the theory of expansion the change in the volume of the object is directly proportional to change in temperature and initial volume.
So here we can say
[tex]\Delta V = V_0\gamma \Delta T[/tex]
here
[tex]\gamma[/tex] = coefficient of volume expansion
so we have
[tex]\gamma = \frac{\Delta V}{V_0 \Delta T}[/tex]
now plug in all values
[tex]\gamma = \frac{2.25 cm^3}{(541 cm^3)(273 - 133)}[/tex]
[tex]\gamma = 2.97 \times 10^{-5} per ^0 C[/tex]
A cyclist is coasting at 12 m/s when she starts down a 450-m-long slope that is 30 m high. The cyclist and her bicycle have a combined mass of 70 kg. A steady 12 N drag force due to air resistance acts on her as she coasts all the way to the bottom. What is her speed at the bottom of the slope?
Answer:
The speed of her at the bottom is 24.035 m/s.
Explanation:
Given that,
Speed of cyclist = 12 m/s
Height = 30 m
Distance d = 450 m
Mass of cyclist and bicycle =70 kg
Drag force = 12 N
We need to calculate the speed at the bottom
Using conservation of energy
K.E+P.E=drag force+K.E+P.E
Potential energy is zero at the bottom.
K.E+P.E=drag force+K.E
[tex]\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}[/tex]
[tex]\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2[/tex]
[tex]25620=5400+35v^2[/tex]
[tex]35v^2=25620-5400[/tex]
[tex]35v^2=20220[/tex]
[tex]v^2=\dfrac{20220}{35}[/tex]
[tex]v=\sqrt{\dfrac{20220}{35}}[/tex]
[tex]v=24.035\ m/s[/tex]
Hence, The speed of her at the bottom is 24.035 m/s.
Answer:
Speed at the bottom of the slope is 24.03 m /sec
Explanation:
We have given speed of the cyclist v = 12 m/sec
She starts down a 450 m long that is 30 m high
Si distance = 450 m and height h = 30 m
Combined mass of cyclist and bicycle m = 70 kg
Drag force = 12 N
We have to find the speed at the bottom
According to conservation theory
KE +PE = work done by drag force + KE + PE
As we know that at the bottom there will be no potential energy
So [tex]\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2[/tex]
[tex]\frac{1}{2}\times 70\times v^2=20220[/tex]
[tex]v=24.03m/sec[/tex]
A projectile launcher is loaded by providing an average force of 23 N to compress a spring 12 cm. If the projectile has a mass of 5.7 grams and is shot at an angle of 57 degrees at a height of 1.37 meters above the floor, what is the spring constant of the launcher, muzzle velocity of the ball, time in the air, maximum height and horizontal distance travelled?
Answer:
a) k = 191.67 N\m
b) V = 22 m/s
c) t = 3.83s
d) 17.36m
e) 45.89 m
Explanation:
given:
F = 23 N
x = 12 cm = 0.12 m
mass of projectile, m = 5.7g = 5.7×10⁻³ kg
a) For a spring
F = kx
where,
F = Applied force
k = spring constant
x = change in in spring length
thus, we have
23 = k×0.12
or
k = 23/0.12
⇒ k = 191.67 N\m
b) From the conservation of energy between the start and the point of interest we have
[tex]\frac{1}{2}\times kx^{2}=\frac{1}{2}mV^{2}[/tex]
where,
V = velocity of the projectile at 1.37 m above the floor
[tex]\frac{1}{2}\times 191.67\times 0.12^{2}=\frac{1}{2}5.7\times 10^{-3}V^{2}[/tex]
V = 22 m/s
c) time in air (t)
applying the Newtons's equaton of motion
[tex]h = Vsin\Theta\times t +\frac{1}{2}a_{y}t^{2}[/tex]
substituting the values in the above equation we get
[tex]-1.37 = 22sin57^{\circ} \times t +\frac{1}{2}(-9.8)t^{2}[/tex]
or
[tex]4.9t^{2}-18.45t-1.37 = 0[/tex]
solving the qudratic equation for 't', we get
t = 3.83s
d) For maximum height ([tex]H_{max}[/tex])
we have from the equations of projectile motion
[tex]H_{max} =\frac{V^{2}sin^{2}\theta }{2g}[/tex]
substituting the values in the above equation we get
[tex]H_{max} =\frac{22^{2}sin^{2}57^{\circ} }{2\times 9.8}[/tex]
or
[tex]H_{max} =17.36 m[/tex]
the height with respect to the ground surface will be = 17.36m + 1.37 m 18.73m
e)For horizontal distance traveled (R) we have the formula
[tex]R = Vcos\theta\times t[/tex]
substituting the values in the above equation we get
[tex]R =22\times cos57^{\circ} \times 3.83[/tex]
or
[tex]R =45.89 m[/tex]
A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the other at 6.8 m/s, 35° north of east. What is the original speed of the mess kit?
Answer:
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Explanation:
Let north represent positive y axis and east represent positive x axis.
Here momentum is conserved.
Let the initial velocity be v.
Initial momentum = 4.4 x v = 4.4v
Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s
Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s
Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s
We have
Initial momentum = Final momentum
4.4v = 12.364 i + 15.092 j
v =2.81 i + 3.43 j
Magnitude
[tex]v=\sqrt{2.81^2+3.43^2}=4.43m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0[/tex]
50.67° north of east.
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance r1 (r1 < R) from the center of the sphere, the electric field has magnitude E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R, the magnitude of the electric field at the same distance r1 from the center would be equal to:
Answer:
[tex]E' = \frac{E}{8}[/tex]
Explanation:
As we know that that electric field inside the solid non conducting sphere is given as
[tex]\int E.dA = \frac{q_{en}}{\epsilon_0}[/tex]
[tex]\int E.dA = \frac{\frac{Q}{R^3}r_1^3}{\epsilon_0}[/tex]
[tex]E(4\pi r_1^2) = \frac{Qr_1^3}{R^3 \epsilon_0}[/tex]
so electric field is given as
[tex]E = \frac{Qr_1}{4\pi \epsilon_0 R^3}[/tex]
now if another sphere has same charge but twice of radius then the electric field at same position is given as
[tex]E' = \frac{Qr_1}{4\pi \epsilon_0 (2R)^3}[/tex]
so here we have
[tex]E' = \frac{E}{8}[/tex]
Final answer:
The electric field at distance r1 from the center of a uniformly charged sphere with charge Q and radius 2R would be one-eighth of the field when the charge is within a sphere with radius R.
Explanation:
To find the electric field at a distance r1 from the center of a uniformly charged sphere, we can use Gauss's law. Inside a uniformly charged sphere, the electric field is proportional to the distance from the center because only the charge enclosed by a Gaussian surface contributes to the field at a point. The field inside the sphere can be described using the formula E = (Qr) / (4πε₀R³), where Q is the total charge, r is the distance from the center, R is the radius of the sphere, and ε0 is the permittivity of free space.
When the same total charge Q is distributed uniformly throughout a sphere of radius 2R, for a point at a distance r1 inside the sphere, which is still less than 2R, the electric field magnitude would be calculated with the new radius. Since the enclosed charge within the distance r1 would be the same and the volume of the larger sphere is greater, the electric field at r1 would be one-eighth of its original value, due to the volume of the sphere increasing by eight times when the radius is doubled (since the volume is proportional to the cube of the radius). Therefore, the new electric field magnitude at distance r1 would be E/8.
Mass m1 is has an initial velocity v1i. Mass m1 collides with a stationary mass m2 . The net external force on the two particle system is zero. If the two masses stick together after the collision, then the final speed vf of the two masses is:
Answer:
vf = (m1*v1i)/(m1 + m2)
Explanation:
take * as multiplication
from the conservation of linear momentum, we know that:
sum of pi = sum of pf, where sum of pi = m1*v1i + m2*v2i
pf = vf(m1 + m2) since the masses stick
together.
v2i = o, since mass m2 is initially stationary.
m1*v1i + m2*(0) = vf(m1 + m2)
m1*v1i = vf(m1 + m2)
therefore
vf = m1*v1i/(m1 + m2)
13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength
The magnetic force acting on a charged particle moving perpendicular to the field is:
[tex]F_{b}[/tex] = qvB
[tex]F_{b}[/tex] is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.
The centripetal force acting on a particle moving in a circular path is:
[tex]F_{c}[/tex] = mv²/r
[tex]F_{c}[/tex] is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.
If the magnetic force is acting as the centripetal force, set [tex]F_{b}[/tex] equal to [tex]F_{c}[/tex] and solve for B:
qvB = mv²/r
B = mv/(qr)
Given values:
m = 1.67×10⁻²⁷kg (proton mass)
v = 7.50×10⁷m/s
q = 1.60×10⁻¹⁹C (proton charge)
r = 0.800m
Plug these values in and solve for B:
B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)
B = 0.979T
The magnetic field strength of the proton is 0.979Tesla
The magnetic force acting on a charged particle moving perpendicular to the field is expressed using the equation.
Fm = qvB
The centripetal force traveled by the proton in a circular path is expressed as:
Fp = mv²/r
To get the field strength, we will equate both the magnetic force and the centripetal force as shown:
Fm = Fp
qvB = mv²/r
qB = mv/r
m is the mass of a proton
v is the velocity = 7.50×10⁷ m/s
m is the mass on the proton = 1.67 × 10⁻²⁷kg
q is the charge on the proton = 1.60×10⁻¹⁹C
r is the radius = 0.800m
Substitute the given parameters into the formula as shown:
[tex]B=\frac{mv}{qr}\\B = \frac{1.67 \times 10^{-27} \times 7.5 \times 10^7}{1.60 \times 10^{-19} \times 0.8}[/tex]
[tex]B=0.979Tesla[/tex]
On solving, the magnetic field strength of the proton is 0.979Tesla
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A cosmic ray proton moving toward the Earth at 5.00×107 m/s experiences a magnetic force of 1.70×10−16 N . What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface?
Explanation:
It is given that,
Speed of proton, [tex]v=5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.7\times 10^{-16}\ N[/tex]
(1) The strength of the magnetic field if there is a 45º angle between it and the proton’s velocity. The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.7\times 10^{-16}\ N}{1.6\times 10^{-19}\ C\times 5\times 10^7\ m/s\ sin(45)}[/tex]
B = 0.00003 T
or
B = 0.03 mT
The magnitude of Earth's magnitude of 25 to 65 Tesla. The value obtained in part (a) is not consistent with the known strength of the Earth’s magnetic field on its surface.
A ball is on the end of a rope that is 1.72 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole's symmetry axis. The rope makes an angle of 64.0° with respect to the vertical. What is the tangential speed of the ball
Answer:
Tangential Speed equals 5.57m/s
Explanation:
In the figure shown for equilibrium along y- axis we have
[tex]\sum F_{y}=0[/tex]
Resolving Forces along y axis we have
[tex]Tcos(\theta )=mg............(i)[/tex]
Similarly along x axis
[tex]\sum F_{x}=ma_{x}[/tex]
[tex]Tsin(\theta )=m[tex]\frac{v^{2} }{r}[/tex]............(ii)[/tex]
Dividing ii by i we have
[tex]tan(\theta )=\frac{v^{2}}{rg}[/tex]
In the figure below we have [tex]r=lsin(\theta )[/tex]
Thus solving for v we have
[tex]v=\sqrt{lgsin(\theta) tan(\theta )}[/tex]
Applying values we get
v=5.576m/s
A tennis player swings her 1000 g racket with a speed of 12 m/s. she hits a 60 g tennis ball that was approaching her at a speed of 16 m/s. The ball rebounds at 42 m/s.
(a) How fast is her racket moving immediatley after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
__________ m/s
(b) If the tennis ball and racket are in contatc for 10 ms, what is the average force that the racket exerts on the ball?
________ N
How does this compare to the gravitational force on the ball?
F avg / W ball=________
Answer:
should be the answer B
Explanation:
A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m. Part A If you suddenly put a 3.0-kg brick in the basket, find the maximum distance that the spring will stretch. (Express your answer in units of cm.)
Answer:
3.92 cm
Explanation:
k = spring constant of the spring scale = 1500 N/m
m = mass of the brick = 3 kg
x = stretch caused in the spring
h = height dropped by the brick = x
Using conservation of energy
Spring potential energy gained by the spring scale = Potential energy lost
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) k x = mg
(0.5) (1500) x = (3) (9.8)
x = 0.0392 m
x = 3.92 cm
Ball A (7kg) and ball B (3kg) hit each other in a perfectly elastic collision. If ball A has an initial velocity of 12 m/s and ball B has an initial velocity of -1 m/s. What are the final velocities of each ball?
Explanation:
Mass of ball A, [tex]m_A=7\ kg[/tex]
Mass of ball B, [tex]m_B=3\ kg[/tex]
Initial velocity of ball A, [tex]u_A=12\ m/s[/tex]
Initial velocity of ball B, [tex]u_B=-1\ m/s[/tex]
We need to find the final velocity of each ball. For a perfectly elastic collision, the coefficient of restitution is equal to 1. It is given by :
[tex]e=\dfrac{v_B-v_A}{u_A-u_B}[/tex]
[tex]v_A\ and\ v_B[/tex] are final velocities of ball A and B
[tex]1=\dfrac{v_B-v_A}{13}[/tex]
[tex]v_B-v_A=13[/tex]...........(1)
Using the conservation of linear momentum as :
[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bu_B[/tex]
[tex]7(12)+3(-1)=7v_A+3u_B[/tex]
[tex]7v_A+3u_B=81[/tex]..............(2)
On solving equation (1) and (2) using calculator we get :
[tex]v_A=4.2\ m/s[/tex]
[tex]v_B=17.2\ m/s[/tex]
So, the final velocities of ball A and B are 4.2 m/s and 17.2 m/s. Hence, this is the required solution.
A ballet dancer has a maximum net torque of 45 Nm, and the moment of inertia of his body is 30 kg mA2. if he starts twirling from rest and keeps accelerating for 95 seconds, how fast will he be rotating? a) b) if he holds rocks in his hands, which variables in this problem would change, in which direction, and why?
Answer:
(a) 142.5 rad/s
Explanation:
Torque = 45 Nm
Moment of inertia = 30 kgm^2
w0 = 0, t = 95 second
(a) Let it is rotating with angular speed w.
Torque = moment of inertia x angular acceleration
45 = 30 x α
α = 1.5 rad/s^2
Use first equation of motion for rotational motion
w = w0 + α t
w = 0 + 1.5 x 95
w = 142.5 rad/s
(b) if he hold rocks, then the moment of inertia change and then angular acceleration change and then final angular velocity change.
Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge Q = -1.85 mC, of mass 5.50 g, is initially located 3.00 cm from both of these charges and released from rest. You observe that the initial acceleration of Q is 314 m>s2 upward, parallel to the line connecting the two point charges. Find q1 and q2.
Answer:
[tex]q_1 = 6.22 \times 10^{-11} C[/tex]
[tex]q_2 = -6.22 \times 10^{-11} [/tex]
Explanation:
Force on the charge Q = -1.85 mC is along the line joining the two charges
so here we can say that net force on it is given by
[tex]F = ma[/tex]
[tex]F = (0.00550)(314)[/tex]
[tex]F = 1.727 N[/tex]
Now this is the force due to two charges which are in same magnitude but opposite sign
so this force is given as
[tex]F = 2\frac{kq_1q}{r^2} cos\theta[/tex]
here we know that
[tex]F = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}cos\theta[/tex]
here we know that
[tex]cos\theta = \frac{2.25}{3} = 0.75[/tex]
now we have
[tex]1.727 = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}(0.75)[/tex]
[tex]1.727 = 2.775 \times 10^{10} q[/tex]
[tex]q = 6.22 \times 10^{-11} C[/tex]
Given that a pipe having a diameter of 1.5 feet and a height of 10 feet, what is the psi at 5 feet ?
Answer:
The pressure is 2.167 psi.
Explanation:
Given that,
Diameter = 1.5 feet
Height = 10 feet
We need to calculate the psi at 5 feet
Using formula of pressure at a depth in a fluid
Suppose the fluid is water.
Then, the pressure is
[tex]P=\rho g h[/tex]
Where, P = pressure
[tex]\rho[/tex] = density
h = height
Put the value into the formula
[tex]P=1000\times9.8\times1.524[/tex]
[tex]P=14935.2\ N/m^2[/tex]
Pressure in psi is
[tex]P=2.166167621\ psi[/tex]
[tex]P=2.167\ psi[/tex]
Hence, The pressure is 2.167 psi.
At one instant, an electron (charge = –1.6 x 10–19 C) is moving in the xy plane, the components of its velocity being vx = 5.0 x 105 m/s and vy = 3.0 x 105 m/s. A magnetic field of 0.80 T is in the positive y direction. At that instant, what is the magnitude of the magnetic force on the electron?
A magnetic field of 0.80 T is there, then the magnitude of the magnetic force on the electron will be equal to - 6.4 x 10⁻¹⁴ k.
What is Charge?Charged material exerts a force when it is introduced to an electromagnetic field because of the chemical property of electric charge. You can have a negative or a positive electric charge. Unlike charge is created one another while like charges repel one another. Neutral refers to an item that carries no net charge.
q = - 1.6 x 10⁻¹⁹ c
v(x) = 5 x 10⁵ m/s,
v(y) = 3 x 10⁵ m/s,
B = 0.8 T along Y-axis
The velocity vector is given by
v = 5 x 10⁵ i + 3 x 10⁵ j
B = 0.8 j
Now calculate the force,
F = q (v x B)
F = -1.6 x 10⁻¹⁹ {(5 x 10⁵ i + 3 x 10⁵ j) x (0.8 j)}
F = - 1.6 x 10⁻¹⁹ (5 x 0.8 x 10^5 k)
F = - 6.4 x 10⁻¹⁴ k
The force's amplitude and direction are along the negative Z axis Is 6.4 x 10⁻¹⁴ N.
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The magnitude of the magnetic force on the electron, given its charge, velocity components, and a magnetic field of 0.80 T in the positive y direction, is 8.0 x 10^(-14) N.
What is the magnitude of the magnetic force on the electron?The magnetic force on a charged particle moving in a magnetic field can be calculated using the following formula:
F = q * v * B * sin(θ)
Where:
- F is the magnetic force.
- q is the charge of the particle.
- v is the velocity of the particle.
- B is the magnetic field strength.
- θ is the angle between the velocity vector and the magnetic field vector.
In this case, the charge of the electron, q, is -1.6 x 10^(-19) C, the velocity components are given as vx = 5.0 x 10^5 m/s and vy = 3.0 x 10^5 m/s, and the magnetic field strength is 0.80 T in the positive y direction.
The angle θ between the velocity and the magnetic field is 90 degrees (π/2 radians) because the electron is moving in the xy plane, and the magnetic field is in the positive y direction.
Now, let's calculate the magnitude of the magnetic force:
F = (-1.6 x 10^(-19) C) * (5.0 x 10^5 m/s) * (0.80 T) * sin(π/2)
F = (-8.0 x 10^(-14) N) * 1
F = -8.0 x 10^(-14) N
So, the magnitude of the magnetic force on the electron is 8.0 x 10^(-14) N.
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For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 ×106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in2 ) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?
Answer:
a) P = 44850 N
b) [tex]\delta l =0.254\ mm[/tex]
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:
[tex]\sigma=\frac{\textup{Load}}{\textup{Area}}[/tex]
on substituting the values, we get
[tex]345\times10^6=\frac{\textup{Load}}{0.00013}[/tex]
or
Load, P = 44850 N
Hence the maximum load that can be applied is 44850 N = 44.85 KN
b)The deformation ([tex]\delta l[/tex]) due to an axial load is given as:
[tex]\delta l =\frac{PL}{AE}[/tex]
on substituting the values, we get
[tex]\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}[/tex]
or
[tex]\delta l =0.254\ mm[/tex]
The maximum load that may be applied to the specimen without causing plastic deformation is 44850 N. The maximum length to which the specimen can be stretched without causing plastic deformation is 76.254 mm.
Explanation:To find the maximum load that may be applied to the specimen without causing plastic deformation, we need to calculate the stress.
Stress = Force / Area
Where the force can be calculated by multiplying the stress with the cross-sectional area of the specimen.
Hence, maximum load = Stress × Area = 345 MPa × 130 mm² = 44850 N
To find the maximum length to which the specimen can be stretched without causing plastic deformation, we need to calculate the strain.
Strain = Change in length / Original length
Maximum length = Original length + (Strain × Original length) = 76 mm + (345 MPa / 103 GPa × 76 mm) = 76 mm + 0.254 mm = 76.254 mm
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For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency (6.8×109 Hz ) and wavelength (0.044 m ). Recall that the Avogadro constant is 6.022×1023 mol−1. Express the energy in joules to two significant figures.
Answer:
2.7 J
Explanation:
The energy of one photon is given by
[tex]E=hf[/tex]
where
h is the Planck constant
f is the frequency
For the photons in this problem,
[tex]f=6.8\cdot 10^9 Hz[/tex]
So the energy of one photon is
[tex]E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J[/tex]
The number of photons contained in 1.0 mol is
[tex]N_A = 6.022\cdot 10^{23} mol^{-1}[/tex] (Avogadro number)
So the total energy of [tex]N_A[/tex] photons contained in 1.0 mol is
[tex]E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J[/tex]
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/sec^2) find the maximum height
Final answer:
To find the maximum height, we need to use the initial velocity and the angle of the baseball's trajectory. By breaking down the initial velocity into its horizontal and vertical components, we can then use the equation h = v_y^2 / (2g) to find the maximum height.
Explanation:
To find the maximum height, we first need to break down the initial velocity into its horizontal and vertical components. The initial velocity of the baseball is given as 100 ft/s at an angle of 45° with respect to the ground.
The horizontal component of velocity can be found using the equation: vx = v * cos(θ), where v is the initial velocity and θ is the angle.
The vertical component of velocity can be found using the equation: vy = v * sin(θ).
Once we have the vertical component of velocity, we can use the equation h = vy2 / (2g) to find the maximum height.
Substituting the given values, we have:
h = (100 * sin(45°))2 / (2 * 32)
Calculating this will give us the maximum height of the baseball above its initial position.
An automobile tire has a radius of 0.344 m, and its center moves forward with a linear speed of v = 20.1 m/s. (a) Determine the angular speed of the wheel. (b) Relative to the axle, what is the tangential speed of a point located 0.135 m from the axle?
Answer:
The angular speed and tangential speed are 58.69 rad/s and 7.92 m/s.
Explanation:
Given that,
Radius = 0.344 m
Speed v= 20.1 m/s
(I). We need to calculate the angular speed
Firstly we will calculate the time
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
[tex]t=\dfrac{2\pi\times r}{v}[/tex]
[tex]t =\dfrac{2\times3.14\times0.344}{20.1}[/tex]
[tex]t=0.107[/tex]
The angular velocity of the tire
[tex]\omega=\dfrac{2\pi}{t}[/tex]
[tex]\omega=\dfrac{2\times3.14}{0.107}[/tex]
[tex]\omega=58.69\ rad/s[/tex]
Now, using formula of angular velocity
(II). We need to calculate the tangential speed of a point located 0.135 m from the axle
The tangential speed
[tex]v = r\omega[/tex]
Where,
r = distance
[tex]\omega[/tex]= angular velocity
Put the value into the formula
[tex]v= 0.135\times58.69[/tex]
[tex]v=7.92\ m/s[/tex]
Hence, The angular speed and tangential speed are 58.69 rad/s and 7.92 m/s.
The position of an object is given by the equation x = 4.0t2 - 2.0 t - 4.5where x is in meters and t is in seconds. What is the instantaneous acceleration, velocity and position of the object at t = 3.0 s? At what time t is the velocity zero. At what time t is the position zero. At what time t is the velocity zero.
Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
[tex]t = \frac{2 \pm \sqrt{4 + 72}}{8}[/tex]
t = 1.4 second
A force of 1.200×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts from rest and is on a level road, what speed ???? will he be going after 25.0 m?
This question requires Newton's Second Law and the formula for kinetic energy to solve. Unfortunately, we lack the man's mass in the problem as posed, making it impossible to determine the velocity. If it is a class question, the student should consult with the professor considering possible typographical errors.
Explanation:This question is about Newton's Second Law of Motion, which states that the acceleration of an object is equal to the net force acting on it divided by the object's mass. In this case, the forces acting on the man are a forward force of 1.200×10³ N and a backwards air resistance force of 615 N. The net force the man feels is therefore (1.200×10³ - 615) N = 585 N.
We also know that the formula for work done (Work = Force x Distance) and that this work is transferred into kinetic energy (Work = 1/2 x Mass x Velocity²). He moved 25.0 m, but we are not given the man's mass to find the velocity directly, making it impossible to solve unless we have the mass. However, if you received this question in class and believe that there is a typo, please consult with your professor for clarification.
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The man will be going approximately 17.1 m/s after 25.0 m.
Sure, here is the solution to the problem:
Given:
Force applied (F_a) = 1.200 × 10³ N
Force resisting (F_r) = 615 N
Distance (d) = 25.0 m
To find:
Final velocity (v)
Solution:
Step 1: Calculate the net force (F_net)
The net force is the difference between the applied force and the resisting force.
F_net = F_a - F_r
F_net = 1.200 × 10³ N - 615 N
F_net = 585 N
Step 2: Calculate the work done (W)
The work done is equal to the net force times the distance.
W = F_net × d
W = 585 N × 25.0 m
W = 14,625 J
Step 3: Calculate the final velocity (v)
The work done is also equal to the change in kinetic energy (ΔKE).
ΔKE = W
(1/2)mv² = 14,625 J
where:
m is the mass of the man and the bicycle
Since the man and the bicycle are initially at rest, their initial velocity (u) is 0. Therefore, the final velocity (v) can be calculated as follows:
v² = 2W / m
v = √(2W / m)
Assuming a mass of 100 kg for the man and the bicycle, we can calculate the final velocity as follows:
v = √((2 × 14,625 J) / (100 kg))
v = √(292.5 J/kg)
v ≈ 17.1 m/s
A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength of the magnetic field if there is a 45° angle between it and the proton's velocity?
Answer:
Magnetic field, [tex]B=4.16\times 10^{-5}\ T[/tex]
Explanation:
It is given that,
Velocity of proton, [tex]v=3.5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.65\times 10^{-16}\ N[/tex]
Charge of proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}[/tex]
B = 0.0000416 T
[tex]B=4.16\times 10^{-5}\ T[/tex]
Hence, this is the required solution.
The strength of the magnetic field, given the force on the proton, its charge, and velocity, and the angle between the velocity and the magnetic field, is approximately 4.21 x 10^-5 T.
Explanation:The strength of a magnetic field can be calculated using the formula for the force exerted on a moving charge in a magnetic field, which is F = q * v * B * sin(θ). Here, F is the magnetic force, q is the charge of the particle, v is the speed of the particle, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field direction.
In this problem, the magnetic force (F) is given as 1.65 x 10^-16 N, the charge of a proton (q) is +1.6 x 10^-19 C, the speed of the proton (v) is 3.5 x 10^7 m/s, and the angle between the velocity and the magnetic field direction (θ) is 45 degrees. Hence we can rearrange the formula to find the magnetic field (B), getting B = F / (q * v * sin(θ)).
Replacing the values into the equation gives: B = 1.65 x 10^-16 N / (+1.6 x 10^-19 C * 3.5 x 10^7 m/s * sin(45°)) which gives the strength of the magnetic field as approximately 4.21 x 10^-5 T.
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A ball dropped from the roof of a tall building will fall approximately how far in two seconds? (A) 10 m, (B) 20 m, (C) 30 m, (D) 40 m, (E) 50 m.
Answer:
The ball covered the distance in 2 second is 20 m.
(b) is correct option.
Explanation:
Given that,
Time = 2 sec
Using equation of motion
[tex]s = ut+\dfrac{1}{2}gt^2[/tex]
Where, s = height
u = initial velocity
g = acceleration due to gravity
t = time
Put the value in the equation
[tex]s = 0+\dfrac{1}{2}\times9.8\times2^2[/tex]
[tex]s = 19.6\ approx\ 20\ m[/tex]
Hence, The ball covered the distance in 2 second is 20 m.
A 360.0 $g$ block is dropped onto a vertical spring with a spring constant k = 254.0 $N/m$. The block becomes attached to the spring, and the spring compresses 0.26 $m$ before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?
Answer:
8.6 J
Explanation:
Work done by the block = change in energy in the spring
W = ½ kx²
W = ½ (254.0 N/m) (0.26 m)²
W = 8.6 J
The windpipe of a typical whooping crane is about 4.6 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 33°C. ans in hz
Answer:
Frequency, f = 481.8 Hz
Explanation:
Given that,
Length of windpipe, l = 4.6 ft = 0.182 m
We need to find the lowest resonant frequency of this pipe at 33 degrees Celcius. Firstly, we will find the speed of sound at 33 degrees Celcius as :
[tex]v=331+0.6T[/tex]
[tex]v=331+0.6\times 33[/tex]
v = 350.8 m/s
At resonance, wavelength is equal to 4 times length of pipe i.e.
λ = 4 l
We need that, [tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f=\dfrac{350.8\ m/s}{4\times 0.182\ m}[/tex]
f = 481.8 Hz
So, the resonant frequency of the windpipe is 481.8 Hz. Hence, this is the required solution.
The lowest resonant frequency of the whooping crane's windpipe, assuming it is closed at one end and a temperature of 33°C, is approximately 62.82 Hz.
Explanation:The lowest resonant frequency of the whooping crane's windpipe, which can be thought of as a pipe that is closed at one end, can be found using the formula for the fundamental frequency of such a pipe: f = v/λ, where f is the frequency, v is the speed of sound, and λ is the wavelength. In this case, the speed of sound is dependent on the temperature and can be approximated as v = 331.4 + 0.6*T m/s, where T is the temperature in degrees Celsius.
At a temperature of 33°C, the speed of sound, v, is approximately 351.8 m/s. Note that we need to convert the length of the windpipe from feet to meters, with 4.6 ft being approximately 1.4 m. The fundamental wavelength for a pipe closed at one end is four times the length of the pipe (λ=4L), so in this case, λ=4*1.4 m = 5.6 m.
Substituting these values into the frequency formula gives us f = v/λ = 351.8 m/s / 5.6 m = 62.82 Hz. So, the lowest resonant frequency of the whooping crane's windpipe is approximately 62.82 Hz.
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An electric field with a magnitude of 6.0 × 104 N/C is directed parallel to the positive y axis. A particle with a charge q = 4.8 × 10–19 C is moving along the x axis with a speed v = 3.0 × 106 m/s. The force on the charge is approximately:
Answer:
Electric force, [tex]F=2.88\times 10^{-14}\ N[/tex]
Explanation:
It is given that,
Magnitude of electric field, [tex]E=6\times 10^4\ N/C[/tex]
Charge, [tex]q=4.8\times 10^{-19}\ C[/tex]
The electric field is directed parallel to the positive y axis. We need to find the force on the charge particle. It is given by :
[tex]F=q\times E[/tex]
[tex]F=4.8\times 10^{-19}\ C\times 6\times 10^4\ N/C[/tex]
[tex]F=2.88\times 10^{-14}\ N[/tex]
So, the electric force on the charge is [tex]2.88\times 10^{-14}\ N[/tex]. Hence, this is the required solution.
A speed skater goes around a turn with a 31 m radius. The skater has a speed of 14 m/s and experiences a centripetal force of 460 N. What is the mass of the skater?
Answer:
Mass of the skater, m = 72.75 kg
Explanation:
It is given that,
Radius of the circular path, r = 31 m
Speed of the skater, v = 14 m/s
Centripetal force, F = 460 N
We need to find the mass of the skater. It can be determined using the formula of centripetal force. It is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]m=\dfrac{Fr}{v^2}[/tex]
[tex]m=\dfrac{460\ N\times 31\ m}{(14\ m/s)^2}[/tex]
m = 72.75 kg
So, the mass of the skater is 72.75 kg. Hence, this is the required solution.
The centripetal acceleration is found to be approximately 6.32 m/s², which helps determine the mass of the skater to be around 72.78 kg.
To find the mass of the speed skater, we will use the formula for centripetal force:
F = m * ac, where:
F is the centripetal force (460 N)
m is the mass of the skater (unknown)
ac is the centripetal acceleration
Centripetal acceleration can be calculated using:
ac = v² / r, where:
v is the speed of the skater (14 m/s)
r is the radius of the turn (31 m)
First, calculate the centripetal acceleration:
ac = (14 m/s)² / 31 m = 196 m²/s² / 31 m ≈ 6.32 m/s²
Now, substitute into the centripetal force formula to solve for the mass (m):
460 N = m * 6.32 m/s²
Solving for m:
m = 460 N / 6.32 m/s² ≈ 72.78 kg
Therefore, the mass of the skater is approximately 72.78 kg.
Suppose you take a trip that covers 1650 km and takes 35 hours to make it. What is your average speed in km/h? Please round your answer to the nearest whole number (integer).
Answer:
47.14 Km/h
Explanation:
distance = 1650 km
time = 35 hours
The average speed is defined as the ratio of total distance to the total time taken.
Average speed = total distance / total time
Average speed = 1650 / 35 = 47.14 Km/h
The container is fitted with a piston so that the volume can change. When the gas is heated at constant pressure, it expands to a volume of 25 L. What is the temperature of the gas in kelvins?
Answer:
[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]
where
[tex]V_{1}[/tex] is initial volume in liters
[tex]T_{1}[/tex] is initial temperature in kelvins
Explanation:
Let the initial volume be [tex]V_{1}[/tex] and the initial temperature be [tex]P_{1}[/tex]
Now by ideal gas law
[tex]\frac{P_{1}V_{1}}{T_{1}}=nR..............(i)[/tex]
Similarly let
[tex]V_{2}[/tex] be final volume
[tex]T_{2}[/tex] be the final volume
thus by ideal gas law we again have
[tex]\frac{P_{2}V_{2}}{T_{2}}=nR..............(ii)[/tex]
Equating i and ii we get
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
For system at constant pressure the above expression reduces to
[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]
Solving for [tex]T_{2}[/tex] we get
[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]
where
[tex]V_{1}[/tex] is initial volume in liters
[tex]T_{1}[/tex] is initial temperature in kelvins
A star has a mass of 1.48 x 1030 kg and is moving in a circular orbit about the center of its galaxy. The radius of the orbit is 1.9 x 104 light-years (1 light-year = 9.5 x 1015 m), and the angular speed of the star is 1.8 x 10-15 rad/s. (a) Determine the tangential speed of the star. (b) What is the magnitude of the net force that acts on the star to keep it moving around the center of the galaxy?
Answer:
a)
3.25 x 10⁵ m/s
b)
8.7 x 10²⁰ N
Explanation:
(a)
w = angular speed of the star = 1.8 x 10⁻¹⁵ rad/s
r = radius of the orbit = 1.9 x 10⁴ ly = 1.9 x 10⁴ (9.5 x 10¹⁵) m = 18.05 x 10¹⁹ m
tangential speed of the star is given as
v = r w
v = (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)
v = 32.5 x 10⁴ m/s
v = 3.25 x 10⁵ m/s
b)
m = mass of the star = 1.48 x 10³⁰ kg
Net force on the star to keep it moving is given as
F = m r w²
F = (1.48 x 10³⁰) (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)²
F = 8.7 x 10²⁰ N